Let $x_1,\dots,x_n$ be i.i.d. drawn from $N(0,I_{p\times p})$. Consider the sample covariance matrix $W(n,p)=\frac 1n \sum_{i=1}^n x_ix_i^T$, a Wishart matrix.

For fixed $n,p$, what is the expected spectral norm of $W(n,p)$, $$\mathbb E \left[ \left\|\frac 1n \sum_{i=1}^n x_ix_i^T\right\|_2\right ] ?$$

I'm familiar with the asymptotic Bai/Silverstein result, but not sure whether that can be exploited in the finite case. If the exact expression is unwieldy, a reasonable upper bound may suffice.

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    $\begingroup$ Cross-posted to stats.stackexchange: stats.stackexchange.com/questions/239916/… Please do not cross-post, especially not after only an hour. Give it a week or so before giving up and posting elsewhere; better would be to flag for migration; at the very least one should notify users, at each site where you post, of all the other SE sites where you post. $\endgroup$
    – Todd Trimble
    Oct 13, 2016 at 13:04

1 Answer 1


I don't think there's an exact expression, but the Bai–Yin result does give the right prediction. It's a little easier to state nice-looking results for a $p \times n$ matrix $X$ with independent standard Gaussian entries, so that what you're asking about is $n^{-1} \mathbb{E} \| X \|_2^2$. The nicest result is that $$ \| X \|_2 \le \sqrt{n} + \sqrt{p} + t $$ with probability at least $1 - 2 e^{-t^2/2}$ (see Theorem II.13 of this paper or Theorem 5.35 of this paper), which implies in particular that $$ \mathbb E \left[ \left\|\frac 1n \sum_{i=1}^n x_ix_i^T\right\|_2\right ] \le \left(1 + \sqrt{\frac{p}{n}}\right)^2 + \frac{4}{n} e^{-(\sqrt{n}+\sqrt{p})^2/2} $$ assuming I did the integrals right. I don't recall offhand a similarly precise result for the lower bound, though there may be one out there. Chances are, though, that the upper bound is what you really need.


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